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That's missing the point, my claim is that correct reason leads us a coherent understanding of some aspects God. This is taught in Islam, where God commands us to use our intellects to arrive at the existence of God, Exalted is He, i.e. "Indeed, in the creation of the heavens and the earth and...

OZYMANDIAS - Percy Bysshe Shelley I met a traveller from an antique land Who said: Two vast and trunkless legs of stone Stand in the desert. Near them on the sand, Half sunk, a shatter'd visage lies, whose frown And wrinkled lip and sneer of cold command Tell that its sculptor well those...

It means 'interpretation' or 'explanation'.

I also don't know why some Muslims claim there are scientific miracles in the Qur'an, it is clearly dodgy exegesis that isn't worthy of the Qur'an, it is exegesis that hasn't appeared in any collections of exegesis, from any sect of Islam The same goes for the "scientific errors", these are...

First of all, this is telling: Reason leads me to belief in a monotheistic Omnipotent God, and it also leads me to a rejection of these central Christian doctrines. Saying that we should just stop using reason is incoherent, there are things are beyond understanding, but blatant...

Re: HSC 2015 4U Marathon good point (idk)

Re: HSC 2015 4U Marathon I 'came up with it' when I was trying to find a way to get 2U students to prove the AM-GM for n=2, but then I realized it could be generalized quite easily haha (with the aid of induction) But I'm sure it's been thought of before

Re: HSC 2015 3U Marathon \\ $i) Explain why$ \ (x^n-1) = (x-1)(1+ x + x^2 + \dots + x^{n-1}) \ $by expanding$ \\ $ii) Hence show that$ \ 1 + x + x^2 + \dots + x^{n-1} = \frac{1-x^{n}}{1-x} \\ $iii) Hence show that$ \ 1 + x + x^2 + \dots = \frac{1}{1-x} \ $for$ \ |x| < 1

Re: HSC 2015 4U Marathon \\ $Consider the polynomial$ \ p(x) = (x+\alpha)^3 - 27\beta x \ $for positive$ \ \alpha, \beta \\\\ $i) Show that if$ \ x \geq 0 \ $and$ \ \alpha^2 \geq 4\beta \ $, then$ \ p(x) \geq 0 \\ $ii) Hence, show that$ \ \frac{a+b+c}{3} \geq \sqrt[3]{abc} \ $for positive$ \...

Re: HSC 2015 4U Marathon I'd add that one way to prove this is by proving the sum is a polynomial (by induction, since if S_n is a polynomial, then S_(n+1) = S_n + (n+1)^2 is a polynomial), and that it's a cubic since the sum is less than: \int_1^n x^2 = \frac{n^3}{3} - \frac{1}{3} (by...

Re: HSC 2015 4U Marathon Alternative to proof given by InteGrand k^2 = 2\binom{k}{2} + \binom{k}{1} \sum_{k=1}^n k^2 = 2\sum_{k=1}^n \binom{k}{2} + \sum_{k=1}^n \binom{k}{1} \\ $Consider the geometric sum$ \\ \\ (1+x) + (1+x)^2 + \dots + (1+x)^n = \frac{1+x}{x} ((1+x)^{n} - 1)...

Re: HSC 2015 4U Marathon - Advanced Level The proof looks solid, well done Here was my approach: \\ $Let$ \ x = b+c-a, y = a+c - b, z = a+b-c \\ $Since$ \ a,b,c \ $are sides of a triangle$ \ x,y,z \ $are all non-negative$ \\ c = \frac{x+y}{2} , b = \frac{x+z}{2} , a = \frac{y+z}{2} \\...

Re: HSC 2015 4U Marathon - Advanced Level nvm

Re: HSC 2015 4U Marathon - Advanced Level \\ a,b,c \ $are sides of a triangle, show that$ \\\\ a^2(b+c-a) + b^2(a+c -b) + c^2(a+b-c) \leq 3abc

Re: HSC 2015 4U Marathon You should mention when the domain is restricted!

Re: HSC 2015 4U Marathon \\ $Show that$ \ \sin^{-1} \left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x \ $for$ \ - 1 \leq x \leq 1

Re: HSC 2015 4U Marathon Fixed that for you :P (and yes that is the approach I was going for) Though for part (i), using the expansion we simply do: \\ \sum_{k=1}^n (k+1)^{m+1} - k^{m+1} = 2^{m+1} - 1^{m+1} + 3^{m+1} - 2^{m+1} + \dots + (n+1)^{m+1} - n^{m+1} = (n+1)^{m+1} - 1 \ (*)...

Re: HSC 2015 4U Marathon The second part of my post was saying that it's not true that S_0(-1) = 0, so that's why I edited in the question to make it "prove for m>=1 that S_m(-1) = 0" (which is true). So, when you sub in n = -1 into the expression in part (i), we get 1 + S_0(-1)...

Re: HSC 2015 2U Marathon

Re: HSC 2015 2U Marathon \\ $Consider the function$ \ f(x) = \tan x - x, \ 0 \leq x < \frac{\pi}{2} \\ \\ $i) Show that$ \ f(x) \ $has exactly one stationary point, and that it is a minimum$ \\ \\ $ii) Hence explain why$ \ f(x) \geq 0 \\ \\ $iii) Hence show that$ \ \frac{1}{\cos x} >...